International American University

Kings College architecting future

Kathmandu, Nepal

Operations Management & Supply Chain: Assignment 1

Submitted by:

Bikram prajapati

ID. No.: 13-S-3043

Submitted to:

Matthew Keogh

 Date of Submission: Dec 1, 2013

Chapter 1

10: Describe each of these system: Craft production, mass production and lean production.

Craft production: craft production is used in earliest days of manufacturing of goods. In craft

production system highly skilled workers use simple flexible tools to produce the goods

according to customer specifications. It doesn’t uses any specialized method to produce the

goods. It is somehow manual process and it is very slow, costly and time consuming. It is used to

produce small amount of goods and also increasing in volume it doesn’t decrease the price of

goods such that there were no economics of scale.

Mass production: After an industrial revolution the companies are started to using the machine

and assemble line to produce huge amount of product. Mass production is system of production

in which low and semi-skilled workers use highly specialized equipment and often costly

machines to produce high volumes of standardize goods. This system is started by Ford

Company. They used highly specialized machine to produce the automobile. It efficiency is very

high and effective than the craft production system and also economies of scale plays an

important role while the goods are produce in huge amount.

Lean production: lean production is Japanese approach to management that focus on reducing

the wastages and it uses minimal amounts of resources i.e. space, machine, inventory and

workers to produce high volume of quality goods. Lean system use a highly skilled workforce

and flexible equipment to produce high quality goods than mass production. The highly skilled

workers will work in maintain and improving system and it tries to cut up cost by making the

business more efficient and responsive to market needs. JIT is one of the lean production system

using by Toyota Company in their system.

Chapter 2

Problem 2

Week Crew size Yards installed Productivity

1 4 960 240

2 3 702 234

3 4 968 242

4 2 500 250

5 3 696 232

6 2 500 250

The productivity is calculated as by the formula:

Productivity=

Yards installedCrewsize

We can see that the productivity is high (250 yard installed) in two case. In both case the

productivity is high when the crew size is 2. So we should chose two people per crew.

Problem 4.

Given, Number of worker = 5

Output = 80 carts per hour, Salary = $ 10 per hour Machine cost= $ 40 per hour

When new system is installed then

Number of worker = 4, Output= 84 carts per hour, Equipment cost= $ 50 per hour

a.) Labor productivity is calculated by Outputlabor

Case Output Workers

Labor productivity

Before 80 5 16After 84 4 21

Productivity changes: 21−16

16∗100 %=31.25 %

i.e. the productivity is increase by 31.25 % after installing the new equipment.

b.) Multifactor productivity is calculated by Output

labor+machine

Case Output Workers

salary machine cost

Multifactor productivity

Before 80 5 50 40 0.89After 84 4 40 50 0.93

Productivity change = 4.49%

The Multifactor Productivity is increase by 4.49 % after installing new equipment.

.

Chapter 3

Problem 1)

Month t Y t*Y

From Table we get n = 7, t = 28, t2 = 140

b=n∑ tY−∑ t∑Y

n∑ t2−(∑ t )2=

7(542 )−28(132)7(140 )−28(28 )

=. 50

a=∑ Y−b∑ t

n=

132−. 50(28 )7

=16 . 86

Feb 1 19 19

Mar 2 18 36

Apr 3 15 45

May 4 20 80

Jun 5 18 90

Jul 6 22 132

Aug 7 20 140

Total 28 132 542

For Sept., t = 8, and

The forecasts sales for September month is Yt = 16.86 + .50(8) = 20.86 units

Feb Mar Apr May Jun Jul Aug0

5

10

15

20

25

Trend Line

Series 1 Linear (Series 1) Linear (Series 1)

2) The forecasts sales by the method of five month moving average is

Time (t) Month Sales (000) 5 month

1 Feb 19  

2 Mar 18  

3 Apr 15  

4 May 20  

5 Jun 18  

6 Jul 22 18

7 Aug 20 18.6

8 Sep 19

I.e.

MA5=15+20+18+22+20

5=19

Unit

3 :) Exponential smoothing technique with smoothing constant = 0.20

Forecast= f (old) + 0.2(actual- f (old))

Time

(t)

Month Sales

(000)

0.2

1 Feb 19 19

2 Mar 18 19

3 Apr 15 18.8

4 May 20 18.04

5 Jun 18 18.432

6 Jul 22 18.3456

7 Aug 20 19.07648

The forecast for September is 19.07 + .2 (20-19.07) = 19.26 Unit

4) The forecast according to naïve method is 20 unit because in naïve method last month sales

will be forecast for upcoming month.

5) The forecast unit by average weighted method is

Time (t) Month Sales (000) Weighted Forecast

1 Feb 19

2 Mar 18

3 Apr 15

4 May 20    

5 Jun 18 0.1  

6 Jul 22 0.3  

7 Aug 20 0.6  

8 Sep     20.4

         

The forecast for September month = 18*0.1 +22*0.3 + 20*0.6 = 20.4 unit

c. Probably 5 month moving average because the data appear to vary around an average of about

19 [18.86].

d. Sales are reflective of demand. To forecast about the future we need to have actual data that

has been sold because it helps to forecast us more accurately

Problem 4.

Week 1 2 3 4 5

Request 20 22 18 21 22

a. Naïve method: the forecast for week 6: 22 requests

b. Four year moving average: the forecast for week 6:

22+18+21+224

=20 .75

Request

c. Exponential smoothing with smoothing constant .30

Given F2= 20 Then

F3 = 20 + .30(22 – 20) = 20.6

F4 = 20.6 + .30(18 – 20.6) = 19.82

F5 = 19.82 + .30(21 – 19.82) = 20.17

F6 = 20.17 + .30(22 – 20.17) = 20.72

The forecast for week 6 is 20.72 request

Problem 6.

As we know the equation of linear trend line is given by: Yt= a + b*t where a= intercept and b=

slope

From graph a= 500 and b= 200/10= 20 Then Y t=500−20 t

Problem 8.

Period t New

account Y

t*Y t2

Now

b=n∑ tY−∑ t∑Y

n∑ t2−(∑ t )2=

15(32076 )−120 (3762)15(1240 )−120 (120 )

=7. 07

a=∑ Y−b∑ t

n=

3762−7 .07 (120 )15

=194 .23

The linear trend equation given by: Yt= 194.23 +

7.07*t

Forecasted demand for periods 16 through 19 are given

by:

Y16 = 194.23 + (7)*(16) = 307.37

Y17 = 194.23 + (7)*(17) = 314.44

Y18 = 194.23 + (7)*(18) = 321.51

Y19 = 194.23 + (7)*(19) = 328.59

b. Initial Trend =

228−2003

=9. 33

1 200 200 1

2 214 428 4

3 211 633 9

4 228 912 16

5 235 1,175 25

6 222 1,332 36

7 248 1,736 49

8 250 2,000 64

9 253 2,277 81

10 267 2,670 100

11 281 3,091 121

12 275 3,300 144

13 280 3,640 169

14 288 4,032 196

15 310 4,650 225

Total

120 3762 32076 1240

Problem 14.

Week

t

Passenger Y t*Y t2

405 405 1

410 820 4

420 1260 9

415 1660 16

412 2060 25

420 2520 36

424 2968 49

433 3464 64

438 3942 81

10 440 4400 100

11 446 4906 121

12 451 5412 144

13 455 5915 169

14 464 6496 196

15 466 6990 225

16 474 7584 256

17 476 8092 289

18 482 8676 324

Total 171 7931 77570 2109

b=(n )(∑ t iY i)−(∑ t i)(∑Y i )

(n )(∑ t i2 )−(∑ t i )

2

b=(18 )(77570)−(171)(7931 )(18 )(2109)−(171)2

=400598721

=4 . 593

a=(∑ Y in )−(b )(∑ ti

n )a=(7931

18 )−( 4 .5933 )(17118 )=396 . 974

The trend line is Yt= 396.974 + 4.593*t

Forecasted demand for the next three weeks are:

Y19 = 396.974 + (4.593)*(19) = 484.25

Y20 = 396.974 + (4.593)*(20) = 488.84

Y21 = 396.974 + (4.593)*(21) = 493.4

Problem 21.

Period Demand F1 e e e2 F2 e e e2

1 68 66 2 2 4 66 2 2 4

2 75 68 7 7 49 68 7 7 49

3 70 72 –2 2 4 70 0 0 0

4 74 71 3 3 9 72 2 2 4

5 69 72 –3 3 9 74 –5 5 25

6 72 70 +

2

2 4 76 –4 4 16

7 80 71 9 9 81 78 2 2 4

8 78 74 4 4 16 80 –2 2 4

32 176 24 106

a. MAD F1: 32/8 = 4.0

MAD F2: 24/8 = 3.0 F2 appears to be more accurate because mean deviation

form trend line is low.

b. MSE F1: 176/7 = 25.14

MSE F2: 106/7 = 15.14 F2 appears to be more accurate because deviation from

mean is low in F2 than F1

Problem 25

a.

X = Price Y = Sales X*Y Y2 X2

6.00 200 1200.00 40,000 36.0000

6.50 190 1235.00 36,100 42.2500

6.75 188 1269.00 35,344 45.5625

7.00 180 1260.00 32,400 49.0000

7.25 170 1232.50 28,900 52.5625

7.50 162 1215.00 26,244 56.2500

8.00 160 1280.00 25,600 64.0000

8.25 155 1278.00 24,025 68.0625

8.50 156 1326.00 24,336 72.2500

8.75 148 1295.00 21,904 76.5625

9.00 140 1260.00 19,600 81.0000

9.25 133 1230.25 17,689 85.5625

Total 92.75 1982 15081 332142 729.06

By calculating we get

b= 19.51 and a= 315.98

The trend line is Y = 315.98 + 19.51 * X

5.5 6 6.5 7 7.5 8 8.5 9 9.50

50

100

150

200

250

Regression Graph

Y Predicted Y

X Variable 1

Y

r=n(∑ xy )−(∑ x )(∑ y )

√n(∑ x2 )−(∑ x )2√n(∑ y2 )−(∑ y )2

= -.9849

b.

r = –.9848. Implies a high, negative relationship between price and demand.

r2 = (–.9848) 2 = .97. It appears that approximately 97% of the variation in sales can be accounted

for by the price of our product. This indicates that price is a good predictor of sales.

Problem 31.

Year Sales(000)

t*y t*t y*y forecast

Error e*e

1 40.2 40.2 1 1616.04 39.79 0.41 0.16812 44.5 89 4 1980.25 43.8 0.7 0.493 48 144 9 2304 47.81 0.19 0.03614 52.3 209.2 16 2735.29 51.82 0.48 0.23045 55.8 279 25 3113.64 55.83 0.03 0.00096 57.1 342.6 36 3260.41 59.84 2.74 7.50767 62.4 436.8 49 3893.76 63.85 1.45 2.10258 69 552 64 4761 67.86 1.14 1.2996

9 73.7 663.3 81 5431.69 71.87 1.83 3.3489Total 503 2756.1 285 29096.08 502.47 8.97 15.1841

By calculating we get b=n∑ tY−∑ t∑Y

n∑ t2−(∑ t )2=35 . 78

And a=

∑ Y−b∑ t

n=4 .01

The regression line is Y= 35.78 + 4.08 *T now the forecast for Next five years are

given as

b.

MSE = 15.15/8 = 1.8938, so s = √1.8938 = 1.376. Control limits are 0 2(1.376) = 0 2.75.

c. The forecast is not in control; two of the last 5 points are outside of the limits.

Problem 33.

A

(sales)

F

(Forecas

t)

A–F

(Error)

Cumulativ

e Error CE

Error Error e2 MA

D

TS=

CE/MAD

15 15 0 0 0 0 0 0 0

Year 10 11 12 13 14

Forecast 75.88 79.89 83.9 87.91 91.92

Year Sales Forecast Error

10 77.2 75.98 1.22

11 82.1 80.00 2.10

12 87.8 84.02 3.78

13 90.6 88.04 2.56

14 98.9 92.05 6.55

21 20 1 1 1 1 1 .5 2

23 25 –2 –1 2 3 4 1 –1

30 30 0 –1 0 3 0 .75 –1.333

32 35 –3 –4 3 6 9 1.2 –3.333

38 40 –2 –6 2 8 4 1

.333

–4.50

42 45 –3 –9 3 11 9 1

.571

–5.729

47 50 –3 –12 3 14 9 1.75 –6.85

MSE=∑ e2

8=36

8−1=5 . 14

Then s=√MSE s=√5. 14=2. 267

No, the forecast is not performing adequately. As you can see from the tracking signal values, we

overestimate the demand consistently. (Tracking signal values continue to get larger.)